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Apr 26, 2015 at 9:50 comment added ubpdqn @FredSimons basically I generate the triples that have the property t1+t2=t3. Next I select from the pairs of triples those that have no common member and make a graph where node/vertex is triple and edge is pair of triples that are disjoint. Originally I tried to build chains but using FindClique finds all mutually disjoint triples (complete subgraph). If there is clique of size n/3 it is a set of mutually disjoint triples that span the range-> a solution.
Apr 25, 2015 at 0:31 comment added ubpdqn @FredSimons thank you for kind words. I make too many mistakes to mention but the most embarrassing ones are also the greatest opportunities, even if mixed with shame...had a lot of shame. I enjoyed thinking about the puzzle and doubtless you and others could find better ways but re looking at small n was useful for me. I am not near computer at present but when time permits will explain approach a bit clearer.
Apr 24, 2015 at 13:04 comment added Fred Simons This is really wonderful! I do not know anything about graph theory (maybe you could very shortly explain the idea behind your solution), but your results show that I must have made at least two errors. One in my proof that for odd nn such partitions do not exist (I found that error, and feel ashamed) and one in my backtracking program that did not find these partitions. That takes more time to find out, but in the very near future I hope to be able to repair it. Many thanks for looking at this problem and congratulations with the result!
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